Physics - Motion

A man at the top of a building 20m high releases a stone from rest; 0.60 seconds later he throws a marble vertically downwards with an initial velocity of 8.0 ms-1. How long after the stone was dropped does the marble pass the stone?

I know that after 0.6 seconds the stone is travelling at 6 ms-1, and because the acceleration is constant, this means that the marble is always travelling 2 ms-1 faster than the stone. I'm thinking that I need to make equations for the distance of each object, equate them and solve for t, but I can't seem to get started. If you could point me in the right direction, that would be great.

Re: Physics - Motion

Originally Posted by Fratricide

A man at the top of a building 20m high releases a stone from rest; 0.60 seconds later he throws a marble vertically downwards with an initial velocity of 8.0 ms-1. How long after the stone was dropped does the marble pass the stone?

I know that after 0.6 seconds the stone is travelling at 6 ms-1, and because the acceleration is constant, this means that the marble is always travelling 2 ms-1 faster than the stone. I'm thinking that I need to make equations for the distance of each object, equate them and solve for t, but I can't seem to get started. If you could point me in the right direction, that would be great.

Re: Physics - Motion

Originally Posted by Fratricide

A man at the top of a building 20m high releases a stone from rest; 0.60 seconds later he throws a marble vertically downwards with an initial velocity of 8.0 ms-1. How long after the stone was dropped does the marble pass the stone?

I know that after 0.6 seconds the stone is travelling at 6 ms-1, and because the acceleration is constant, this means that the marble is always travelling 2 ms-1 faster than the stone.

Are you rounding to one decimal place? Why would you do that? The downward acceleration, due to gravity, is 9.2 m/s2 so that after 0.6 seconds, the stone will be travelling at (9.2)(0.6)= 5.52 m/s downward. So the difference in speeds will be 8.00- 5.52= 2.48 m/s. In any case, the height of the stone, t seconds after it was thrown will be -4.6t^2+ 20 while the height of the marble will be -4.6(t- .6)^2- 8(t- .6)+ 20. The marble will "pass" the stone where their heights are equal: -4.6t^2+ 20= -4.6(t- .6)^2- 8(t- .6)+ 20. The t^2 terms will cancel out leaving a linear equation to solve for t.

I'm thinking that I need to make equations for the distance of each object, equate them and solve for t, but I can't seem to get started. If you could point me in the right direction, that would be great.

Re: Physics - Motion

All of my textbooks, assignments and exams for the course recognise the acceleration due to gravity as 10m/s2 (which obviously isn't accurate), and as such my physics teacher asks us to use that value in our calculations.

Re: Physics - Motion

Originally Posted by Fratricide

All of my textbooks, assignments and exams for the course recognise the acceleration due to gravity as 10m/s2 (which obviously isn't accurate), and as such my physics teacher asks us to use that value in our calculations.